x4+1x4−2x^4 + \dfrac{1}{x^4} - 2x4+x41−2 in the form of factors is :
(x−1x)2(x+1x+1)2\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} + 1\Big)^2(x−x1)2(x+x1+1)2
(x−1x)2(x+1x)2\Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2(x−x1)2(x+x1)2
(x+1x)2(x−1x+1)2\Big(x + \dfrac{1}{x}\Big)^2\Big(x - \dfrac{1}{x} + 1\Big)^2(x+x1)2(x−x1+1)2
(x+1x)2(x+1x−1)2\Big(x + \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x} - 1\Big)^2(x+x1)2(x+x1−1)2
10 Likes
Given,
=x4+1x4−2=(x2)2+(1x2)2−2×x2×1x2=(x2−1x2)2=[(x−1x)(x+1x)]2=(x−1x)2(x+1x)2.\phantom{=}x^4 + \dfrac{1}{x^4} - 2 \\[1em] = (x^2)^2 + \Big(\dfrac{1}{x^2}\Big)^2 - 2 \times x^2 \times \dfrac{1}{x^2} \\[1em] = \Big(x^2 - \dfrac{1}{x^2}\Big)^2 \\[1em] = \Big[\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)\Big]^2 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2\Big(x + \dfrac{1}{x}\Big)^2.=x4+x41−2=(x2)2+(x21)2−2×x2×x21=(x2−x21)2=[(x−x1)(x+x1)]2=(x−x1)2(x+x1)2.
Hence, Option 2 is the correct option.
Answered By
4 Likes
Factorise :
8a3 - b3 - 4ax + 2bx
a - b - a3 + b3
x2+1x2−3x^2 + \dfrac{1}{x^2} - 3x2+x21−3 in the form of factors is :
(x−1x)(x+1x)\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)(x−x1)(x+x1) - 1
(x+1x+1)(x−1x−1)\Big(x + \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x+x1+1)(x−x1−1)
(x−1x)(x−1x−1)\Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x−x1)(x−x1−1)
(x−1x+1)(x−1x−1)\Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x−x1+1)(x−x1−1)
x2+14x2+1−7x−72xx^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x}x2+4x21+1−7x−2x7
